What is (1+2i)(1-2i)(3+4i) ?

The solution to (1+2i)(1-2i)(3+4i) is 15+20i

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simplify (1+2i)(1-2i)(3+4i)

simplify

(1 + 2 i) (1 - 2 i) (3 + 4 i)

15 + 20 i

Solution steps

(1 + 2 i) (1 - 2 i) (3 + 4 i)

Apply complex arithmetic rule:

(a + bi) (a - bi) = a^{2} + b^{2}

= (1^{2} + 2^{2}) (3 + 4 i)

Simplify

(1^{2} + 2^{2}) (3 + 4 i) : 15 + 20 i

= 15 + 20 i

20 x^{5}

32 \cdot 24

(x^{2} + 5) (x^{2} + 5)

- 325 - 253

\frac{75}{7}